Binomial Sum
Inviato: 06 ago 2009, 19:24
Evaluate $ \displaystyle \sum_{n} x^n \binom{2n}{n}\binom{l}{n} $ where n and l are natural numbers
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Binomial Sum
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Inviato: 06 ago 2009, 20:01
gosh.. maybe I forgot the original question: it cannot be this!
Inviato: 06 ago 2009, 21:38
You forgot way more than that.
sonata
Inviato: 11 ago 2009, 20:13
Ah!
I must repair someway. But I cannot delucidate here where this stupid question originated, it would be tremendously OT.
By the way, there is no path now to come out from this vacuum.
I'll try to hypnotize you with this, in the hope that nobody already knows it.
Then listen to that player.
Well; now you are in my power! therefore I can explain you the whys of the question without you being able to recognise the OT situation.
Say yes with your head
good job, dear.
Right, the sum comes out if you try to determine in the path integral perturbation theory the eigenvalues of the hamiltonian with a potential $ \frac{1}{2}(\omega^2q^2+\lambda^2 q^2) $ as a function of $ \lambda $. I never used such tools, so I believed it would give me the trivial exact result; three years of physics built such a confusion about simple mathematics that I thought the sum should have the expected simple result.. here is the argument (page 3).
Did you hear the wrong note? Poor Pius
I must repair someway. But I cannot delucidate here where this stupid question originated, it would be tremendously OT.
By the way, there is no path now to come out from this vacuum.
I'll try to hypnotize you with this, in the hope that nobody already knows it.
Then listen to that player.
Well; now you are in my power! therefore I can explain you the whys of the question without you being able to recognise the OT situation.
Say yes with your head
good job, dear.
Right, the sum comes out if you try to determine in the path integral perturbation theory the eigenvalues of the hamiltonian with a potential $ \frac{1}{2}(\omega^2q^2+\lambda^2 q^2) $ as a function of $ \lambda $. I never used such tools, so I believed it would give me the trivial exact result; three years of physics built such a confusion about simple mathematics that I thought the sum should have the expected simple result.. here is the argument (page 3).
Did you hear the wrong note? Poor Pius
Inviato: 12 ago 2009, 02:47
It's not cool to do maths when you're high on crack. 
Inviato: 12 ago 2009, 15:15
Tibor Gallai ha scritto:It's cool to do maths when you're high on crack.
Oh.. How I agree babe!
Inviato: 12 ago 2009, 15:26
As I don't see nor a sensible mathematical question neither any semblance of seriousness in this thread, I'll close it.
I warmly invite you, dear maxpower, to stop posting such nonsenses; however, if you want to participate actively in our discussions, you'll be mostly welcome (even if the language used in this forum usually is Italian).
I warmly invite you, dear maxpower, to stop posting such nonsenses; however, if you want to participate actively in our discussions, you'll be mostly welcome (even if the language used in this forum usually is Italian).